Question

The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis and the line AB at C and E respectively. If O is the origin of coordinates, find the area of figure OCEB.

Answer 1

The given line is 2x + 3y = 12, which can be written as

$\frac{x}{6}+\frac{y}{4}=1$ ... (1)

So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively.



The equation of the line perpendicular to line (1) is

$\frac{x}{4}-\frac{y}{6}+\lambda =0$

This line passes through the point (5, 5).

$$\begin{gathered} \therefore \frac{5}{4}-\frac{5}{6}+\lambda =0 \\ \Rightarrow \lambda =-\frac{5}{12} \end{gathered}$$

Now, substituting the value of $\lambda$ in $\frac{x}{4}-\frac{y}{6}+\lambda =0$, we get:
$$\begin{gathered} \frac{x}{4}-\frac{y}{6}-\frac{5}{12}=0 \\ \Rightarrow \frac{x}{{\displaystyle \frac{5}{3}}}-\frac{y}{{\displaystyle \frac{5}{2}}}=1...\left(2\right) \end{gathered}$$


Thus, the coordinates of intersection of line (1) with the x-axis is $C\left(\frac{5}{3},0\right)$.

To find the coordinates of E, let us write down equations (1) and (2) in the following manner:

$2x+3y-12=0$ ... (3)

$3x-2y-5=0$ ... (4)

Solving (3) and (4) by cross multiplication, we get:

$$\begin{gathered} \frac{x}{-15-24}=\frac{y}{-36+10}=\frac{1}{-4-9} \\ \Rightarrow x=3,y=2 \end{gathered}$$

Thus, the coordinates of E are (3, 2).

From the figure,

$EC=\sqrt{{\left(\frac{5}{3}-3\right)}^2+{\left(0-2\right)}^2}=\frac{2\sqrt{13}}{3}$



$EA=\sqrt{{\left(6-3\right)}^2+{\left(0-2\right)}^2}=\sqrt{13}$

Now,

$$\begin{gathered} \mathrm{Area}\left(OCEB\right)=\mathrm{Area}\left(∆OAB\right)-\mathrm{Area}\left(∆CAE\right) \\ \Rightarrow \mathrm{Area}\left(OCEB\right)=\frac{1}{2}\times 6\times 4-\frac{1}{2}\times \frac{2\sqrt{13}}{3}\times \sqrt{13} \\ =\frac{23}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$